Is ${919872}$ divisible by $9$ ?
Solution: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {919872}= &&{9}\cdot100000+ \\&&{1}\cdot10000+ \\&&{9}\cdot1000+ \\&&{8}\cdot100+ \\&&{7}\cdot10+ \\&&{2}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {919872}= &&{9}(99999+1)+ \\&&{1}(9999+1)+ \\&&{9}(999+1)+ \\&&{8}(99+1)+ \\&&{7}(9+1)+ \\&&{2} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {919872}= &&\gray{9\cdot99999}+ \\&&\gray{1\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {9}+{1}+{9}+{8}+{7}+{2} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${919872}$ is divisible by $9$ if ${ 9}+{1}+{9}+{8}+{7}+{2}$ is divisible by $9$ Add the digits of ${919872}$ $ {9}+{1}+{9}+{8}+{7}+{2} = {36} $ If ${36}$ is divisible by $9$ , then ${919872}$ must also be divisible by $9$ ${36}$ is divisible by $9$, therefore ${919872}$ must also be divisible by $9$.